The end B of the rod AB which makes angle the | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$V_{A} \cos (90-	heta)=V_{0} \cos 	heta$

$\mathrm{V}_{\mathrm{A}} \sin 	heta=\mathrm{V}_{0} \cos 	heta$

$\mathrm{V}_{\mathrm{A}}=\frac{\math

Vedclass · Accepted Answer

angular velocity of rod is $\frac{5}{3} \frac{v_0}{l}$

Vedclass · Answer

$V_{A} \cos (90-	heta)=V_{0} \cos 	heta$

$\mathrm{V}_{\mathrm{A}} \sin 	heta=\mathrm{V}_{0} \cos 	heta$

$\mathrm{V}_{\mathrm{A}}=\frac{\mathrm{V}_{0}}{	an 	heta}$

$\mathrm{V}_{\mathrm{A}}=\frac{4}{3} \mathrm{V}_{0}$

$\omega=\frac{\left[V_{A} \sin (90-	heta)+V_{0} \sin 	hetaight]}{\ell}$

The end $B$ of the rod $AB$ which makes angle $\theta$ with the floor is being pulled with, a constant velocity $v_0$ as shown. The length of the rod is $l.$ At the instant when $\theta = 37^o $ then

Similar Questions

An elevator accelerates upwards at a constant rate. A uniform string of length $L$ and mass $m$ supports a small block of mass $M$ that hangs from the ceiling of the elevator. The tension at distance $l$ from the ceiling is $T$ . The acceleration of the elevator is

Find the velocity of the hanging block if the velocities of the free ends of the rope are as indicated in the figure.

In the arrangement shown in fig. the ends $P$ and $Q$ of an unstretchable string move downwards with uniform speed $U$. Pulleys $A$ and $B$ are fixed. Mass $M$ moves upwards with a speed.

Find velocity of block ' $B$ ' at the instant shown in figure $........\,m/s$

Two particles $A$ and $B$ are connected by rigid rod $A B$. The rod slides along perpendicular rails as shown here. The velocity of $A$ to the left is $10\; m / s$. What is the velocity of $B$(in $m/s$) when angle $\alpha=60^{\circ}$?